If $f(x) = \frac{x^2}{2 - 2 \cos x}$; $g(x) = \frac{x^2}{6x - 6 \sin x}$ where $0 < x < 1$, then (A) 'f' is increasing function (B) 'g' is decreasing function (C) $\frac{f(x)}{g(x)}$ is increasing function (D) $g(f(x))$ is decreasing function

1. Why study increasing or decreasing?

When a function is increasing on an interval, every time you move the input a little to the right, the output also moves up. If it is decreasing, the output moves down instead. The usual tool is the first derivative:

• If $f'(x) > 0$ on an interval → $f$ is increasing there.
• If $f'(x) < 0$ on an interval → $f$ is decreasing there.

2. Function $f(x) = \dfrac{x^2}{2 - 2\cos x}$

Rewrite the denominator using the identity $1 - \cos x = 2\sin^2 \dfrac{x}{2}$:

$2 - 2\cos x = 4\sin^2\frac{x}{2}$

So

$f(x) = \left(\frac{x}{2\sin\frac{x}{2}}\right)^2.$

Call the simpler piece $h(x) = \dfrac{x}{\sin\dfrac{x}{2}}$. Because the square function keeps the same ordering for positive numbers, $f$ is increasing exactly when $h$ is increasing.

Differentiate $h$:

$h'(x) = \frac{2\sin\frac{x}{2} - x\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}.$

On $(0,1)$ both the numerator and denominator are positive (a short series‐expansion check or plugging a few points confirms this), so $h'(x) > 0$ and hence $f$ is increasing.

3. Function $g(x) = \dfrac{x^2}{6x - 6\sin x}$

Factor the constant 6:

$g(x)=\frac{x^2}{6(x-\sin x)} = \tfrac16\;\underbrace{\left(\frac{x^2}{x-\sin x}\right)}_{k(x)}.$

The constant $\tfrac16$ cannot change monotonicity, so investigate

$k(x) = \frac{x^2}{x-\sin x}.$
Take natural logs to avoid messy quotients:

$\ln k = 2\ln x - \ln\bigl(x - \sin x\bigr).$

Differentiate:

$\frac{k'}{k} = \frac{2}{x} - \frac{1-\cos x}{x-\sin x}.$

For $x\in(0,1)$ one finds the second term is always larger than the first, making the entire derivative negative. Therefore $g$ is decreasing.

4. Ratio $\dfrac{f(x)}{g(x)}$

Cancelling common $x^2$ factors gives

$\frac{f(x)}{g(x)} = \frac{6x-6\sin x}{2-2\cos x} = 3\,\frac{x-\sin x}{1-\cos x}=3\,r(x).$

Compute the derivative of $r(x)$ and show it is positive (small‐$x$ series and a couple of spot checks on the interval confirm this). Hence the ratio is increasing.

5. Composition $g\bigl(f(x)\bigr)$

Since $f$ is increasing and $g$ is decreasing, the composition $g\circ f$ is decreasing: as $x$ grows, $f(x)$ grows, but a decreasing $g$ flips the direction.

Bottom line

All four statements (A), (B), (C), (D) are correct on the interval $(0,1)$.

Imagine two magic boxes

1. Box f takes a tiny positive number (between 0 and 1), jiggles it with a special rule, and gives a new number.
2. Box g does something similar but with a slightly different rule.

We are asked four questions:

1. Does the output of box f always get bigger if we feed a bigger input?
2. Does the output of box g always get smaller if we feed a bigger input?
3. If we compare what the two boxes spit out, does the ratio grow when we feed larger inputs?
4. If we first use box f and then feed its result into box g, does the final answer always go down when the first input goes up?

The short, friendly answer is: Yes, Yes, Yes, and Yes!

So, all four statements (A), (B), (C), (D) are true.

Step-by-step calculations

1. Rewrite $f$.

   $f(x)=\frac{x^2}{2-2\cos x}=\left(\frac{x}{2\sin \dfrac{x}{2}}\right)^2.$

2. Derivative test for $f$.

   $h(x)=\frac{x}{\sin\tfrac{x}{2}},\quad f\text{ increasing }\iff h'(x)>0.$

   $h'(x)=\frac{2\sin\tfrac{x}{2}-x\cos\tfrac{x}{2}}{2\sin^2\tfrac{x}{2}}>0\quad (0<x<1).$

   ⇒ $f$ is increasing. (A ✓)

3. Derivative test for $g$.

   $g(x)=\frac{x^2}{6(x-\sin x)}\equiv\tfrac16\,k(x),\quad k(x)=\frac{x^2}{x-\sin x}.$

   $\frac{k'}{k}=\frac{2}{x}-\frac{1-\cos x}{x-\sin x}<0 \;(0<x<1)\;\Rightarrow\;g'(x)<0.$

   ⇒ $g$ is decreasing. (B ✓)

4. Ratio $f/g$.

   $\frac{f(x)}{g(x)}=3\,\frac{x-\sin x}{1-\cos x}=3\,r(x).$

   $r'(x)=\frac{(1-\cos x)^2-(x-\sin x)\sin x}{(1-\cos x)^2}>0\;(0<x<1)$

   (both series expansion and numerical checks confirm positivity) ⇒ ratio increasing. (C ✓)

5. Composition $g(f(x))$.

   • $f$ ↑ (increasing)
   • $g$ ↓ (decreasing)

   Increasing input to a decreasing function makes the whole chain decreasing ⇒ (D ✓)

Final answer:
$(A), (B), (C), (D)$ are all correct.